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Author  Topic

Heather
Junior Member

USA
4 Posts

 Posted - 10/23/2007 :  19:37:51 I can't figure out this identity, I was wondering if someone could help me with it... sin-tancos-cot=tan^6

skeeter

USA
5634 Posts

 Posted - 10/23/2007 :  19:49:14 (sinx - tanx)/(cosx - cotx) =[sinx(1 - secx)]/[cosx(1 - cscx)] =(sinx)/(cosx)*(1 - secx)/(1 - cscx) =tanx * (-tanx)/(-cotx) =tanx * tanx * tanx = tan6x

Heather
Junior Member

USA
4 Posts

 Posted - 10/23/2007 :  20:16:45 Thank you sooo much!! it was like a big DUH when i saw it! haha thanks again!
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